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Phonics is an instructional method in which children are taught to connect sounds with letters or groups of letters. A sample of 134 first-graders who were learning English were asked to identify as many letter sounds as possible in a period of one minute. The average number of letter sounds identified was 34.06 with a standard deviation of 23.83.

a. Construct a 98% confidence interval for the mean number of letter sounds identified in one minute.
b. If a 95% confidence interval were constructed with these data, would it be wider or narrower than the interval constructed in part (a)? Explain.

Answer :

Answer:

a) [tex]34.06-2.355\frac{23.83}{\sqrt{134}}=29.212[/tex]    

[tex]34.06+2.355\frac{23.83}{\sqrt{134}}=38.908[/tex]    

b) For this case since the confidence level decrease then the confidence interval would decrease respect a) also since the margin of error would be lower.

Step-by-step explanation:

Information given

[tex]\bar X=34.06[/tex] represent the sample mean

[tex]\mu[/tex] population mean (variable of interest)

s=23.83 represent the sample standard deviation

n=134 represent the sample size  

Part a

The confidence interval for the mean is given by the following formula:

[tex]\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}[/tex]   (1)

The degrees of freedom are given:

[tex]df=n-1=134-1=133[/tex]

The Confidence level is 0.98 or 98%, the value of significance is [tex]\alpha=0.02[/tex] and [tex]\alpha/2 =0.01[/tex],  and the critical value would be [tex]t_{\alpha/2}=2.355[/tex]

Replacing we got:

[tex]34.06-2.355\frac{23.83}{\sqrt{134}}=29.212[/tex]    

[tex]34.06+2.355\frac{23.83}{\sqrt{134}}=38.908[/tex]    

Part b

For this case since the confidence level decrease then the confidence interval would decrease respect a) also since the margin of error would be lower.

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